## University Calculus: Early Transcendentals (3rd Edition)

$g(x)=\sqrt[3]{\frac{x+6}{2}}$
$f(x)=2x^3-4$ $(f\circ g)(x)=x+2$ $g(x)=y=?$ The composition of the two functions is represented as: $(f\circ g)(x)=f(g(x))$ $f(y)=2(y)^3-4$ $x+2=2y^3-4$ Solve this equation for $\ y\$ to get: $x+6=2y^3$ $y^3=\frac{x+6}{2}$ $y=\sqrt[3]{\frac{x+6}{2}}$ $g(x)=\sqrt[3]{\frac{x+6}{2}}$